How to access the previous/next element in a for loop?

Expressed as a generator function:

def neighborhood(iterable):
    iterator = iter(iterable)
    prev_item = None
    current_item = next(iterator)  # throws StopIteration if empty.
    for next_item in iterator:
        yield (prev_item, current_item, next_item)
        prev_item = current_item
        current_item = next_item
    yield (prev_item, current_item, None)

Usage:

for prev,item,next in neighborhood(l):
    print prev, item, next

l = [1, 2, 3]
for i, item in enumerate(l):
    if item == 2:
        previous = l[i - 1]
        print(previous)

Output:

1

This will wrap around and return the last item in the list if the item you're looking for is the first item in list. In other words changing the third line to if item == 1: in the above code will cause it to print 3.

When dealing with generators where you need some context, I often use the below utility function to give a sliding window view on an iterator:

import collections, itertools

def window(it, winsize, step=1):
    """Sliding window iterator."""
    it=iter(it)  # Ensure we have an iterator
    l=collections.deque(itertools.islice(it, winsize))
    while 1:  # Continue till StopIteration gets raised.
        yield tuple(l)
        for i in range(step):
            l.append(it.next())
            l.popleft()

It'll generate a view of the sequence N items at a time, shifting step places over. eg.

>>> list(window([1,2,3,4,5],3))
[(1, 2, 3), (2, 3, 4), (3, 4, 5)]

When using in lookahead/behind situations where you also need to deal with numbers without having a next or previous value, you may want pad the sequence with an appropriate value such as None.

l= range(10)
# Print adjacent numbers
for cur, next in window(l + [None] ,2):
    if next is None: print "%d is the last number." % cur
    else: print "%d is followed by %d" % (cur,next)

Check out the looper utility from the Tempita project. It gives you a wrapper object around the loop item that provides properties such as previous, next, first, last etc.

Take a look at the source code for the looper class, it is quite simple. There are other such loop helpers out there, but I cannot remember any others right now.

Example:

> easy_install Tempita
> python
>>> from tempita import looper
>>> for loop, i in looper([1, 2, 3]):
...     print loop.previous, loop.item, loop.index, loop.next, loop.first, loop.last, loop.length, loop.odd, loop.even
... 
None 1 0 2 True False 3 True 0
1 2 1 3 False False 3 False 1
2 3 2 None False True 3 True 0

If you don't want to import anything, here is an example of accessing the previous item of a generator with the for loop. It uses a class variable to store each next result before the following next call. This variable could be a small list if you wanted more than the immediately previous item. Inside the class is a method generator that effectively extends the next() builtin to include the previous item assignment.

Code (Python 3.10):

def previous():
    class Plusprev():
        def __init__(pp, gen=None):
            pp.g = gen
            pp.nxt = ''
            pp.prev = 'start'

        def ppnext(pp):
            while pp.nxt != 'done':
                pp.nxt = next(pp.g,'done')
                yield pp.nxt
                pp.prev = pp.nxt

    sqgen = (n*n for n in range(13))
    ppcl = Plusprev(sqgen)
    nxtg = ppcl.ppnext()
    nxt = next(nxtg,'done')
    while nxt != 'done':
        print('\nprevious ',ppcl.prev)
        print('current ',nxt)
        nxt = next(nxtg,'done')

previous()

This uses the builtin function, next(), default parameter.

I don't think there is a straightforward way, especially because an iterable can be a generator (no going back). You can do it with sequences by passing the index of the element into the loop body:

for index, item in enumerate(l):
    if index > 0:
        previous_item = l[index - 1]
    else:
        previous_item = None 

The enumerate() function is a builtin.

Immediately previous?

You mean the following, right?

previous = None
for item in someList:
    if item == target: break
    previous = item
# previous is the item before the target

If you want n previous items, you can do this with a kind of circular queue of size n.

queue = []
for item in someList:
    if item == target: break
    queue .append( item )
    if len(queue ) > n: queue .pop(0)
if len(queue ) < n: previous = None
previous = previous[0]
# previous is *n* before the target

I know this is an old question but I find it important to show a simple solution which works also with generators and other kind of iterables, not like most of the answers working only with list like objects. It is somewhat similar to Brian's answer and the solution here: https://www.programcreek.com/python/example/1754/itertools.tee

import itertools

iter0, iter1 = itertools.tee(iterable)

for item, next_item in itertools.zip_longest(
    iter0,
    itertools.islice(iter1, 1, None)
):

    do_something(item, next_item)

Alternatively, calling next on the second iterable (if you are sure it has at least one element):

import itertools

iter0, iter1 = itertools.tee(iterable)
_ = next(iter1)

for item, next_item in itertools.zip_longest(iter0, iter1):

    do_something(item, next_item)

For anyone who upgraded to python 3.10, such function was added directly to itertools

import itertools

l = [1,2,3]
for x, y in itertools.pairwise(l):
    print(x, y)
# 1 2
# 2 3

Iterators only have the next() method so you cannot look forwards or backwards, you can only get the next item.

enumerate(iterable) can be useful if you are iterating a list or tuple.

Not very pythonic, but gets it done and is simple:

l=[1,2,3]
for index in range(len(l)):
    if l[index]==2:
        l[index-1]

TO DO: protect the edges

The most simple way is to search the list for the item:

def get_previous(l, item):
    idx = l.find(item)
    return None if idx == 0 else l[idx-1]

Of course, this only works if the list only contains unique items. The other solution is:

for idx in range(len(l)):
    item = l[idx]
    if item == 2:
        l[idx-1]